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WHAT WOULD HAPPEN IF EVERYONE JUMPED??

If everyone in the world (7.6 billion people) gathered in one place and all jumped at the same time, what would happen to the Earth? Would it be effected? 


This is a question that scientists have been asking for years. The answer simply put is that even if everyone in the world jumped at once, the Earth would virtually be unaffected. This is because the mass of the earth is approximately 5.972 x 10^24 kg, and all human mass put together would not even come close to that number. To see the exact number of how the Earth would be effected, I used the conservation of momentum equation. 

p1i + p2i = p1f + p2f
m1v1+m2v2=m1v1f+m2v2f

I know that in this situation the equation is elastic because the Earth is in the system and momentum is always conserved. In all my calculations, 1 will represent the Earth and 2 will represent the people. 

First here are the numbers I used for the data:
  • Average human mass = 136 lbs / 62 kg (<https://www.livescience.com/36470-human-population-weight.html>)
  • Human population as of December 2017 = 7.6 billion
  • Mass of the Earth = 5.972 x 10^24 kg 
  • Average vertical jump = 0.5m
Because this is an elastic collision, this means that energy is also conserved. I started this problem using a conservation of energy equation to solve for the velocity of the Earth. This I showed by just using the initial point before the jump when delta KE = delta PE. The Earth did not move up or down, so h = 0.

1/2m1v1^2 + 1/2m2v2^2 = mgh
h initial = 0 
1/2m1v1^2 + 1/2m2v2^2 = 0
1/2m1v1^2 = -1/2m2v2^2
m1v1= - m2v2
v1 = -m2v2/m1

I then used the same energy equation to calculate the velocity of the people, when h did not = 0. The mass in the potential energy is the m2: the mass of the people. 


1/2m1v1^2 + 1/2m2v2^2 = m2gh
m1v1^2 + m2v2^2 = 2m2gh
m2v2^2 = 2m2gh-m1v1^2
v2^2 = (2m2gh-m1v1^2)/m2
v2^2 = 2gh - m1v1^2/m2
v2 = sq. rt. 2gh - m1v1^2/m2

I then put the two equations together to solve for the velocity of the Earth. Here was the result:

v1^2 = (2ghm2^2/m1^2) - (m2v1^2/m1)

From here I had to solve for v1 to get the velocity of the Earth.

v1^2 + (m2v1^2/m1) = (2ghm2^2/m1^2)
v1^2 (1 +m2/m1) = (2ghm2^2/m1^2)
v1^2 = 2ghm2^2 / m1^2 (1 +m2/m1)
v1^2 = 2ghm2^2/ (m1^2+m2m1)
v1 = sq. rt. 2ghm2^2/ (m1^2+m2m1)

Now that I found this equation, all I had to do was plug in the values for each thing that I mentioned above. 
First I found the mass of the total population of the people which was (7.6 x 10^9)*62 = 4.712 x 10^11 kg. 

v2 = sq. rt. 2*9.8* (4.712 x 10^11)^2 / (5.972 x 10^24)^2 + (5.972 x 10^24)*(4.712 x 10^11)
v2 = (2.17589 x 10^24)/ (3.56648 x 10^49)
v2 = 6.10095 x 10^-26 m/s

This means that the velocity of the Earth after the jump would only be effected by 6.10095 x 10^-26 m/s. Virtually, there would be no effect on the Earth. This is probably because the total mass of the human population is so small compared to the Earth it would be hard to effect the Earth's momentum. The larger the mass, the harder it is to move. This means in order for the human population to change the Earth's orbit they would need a giant number for their velocity in order to make a dent. 

So at the end of the day, if everyone decided to gather in the same place and jump it would probably just be a waste of their time. 

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